Gujarat Board GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Question 1.

Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2: 3.

Solution:

Let the given points be A(-1, 7) and B(4, -3).

Here, we have

x_{1} = -1, y_{1} = 7

x_{2} = 4, y_{2} = -3

and m_{1} = 2, m_{2} = 3

Let the required point be P(x, y).

Then, x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)

and y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)

⇒ x = \(\frac{2(4)+3(-1)}{2+3}\) = \(\frac{8-3}{5}\) = \(\frac{5}{5}\) = 1

And y = \(\frac{2(-3)+3(7)}{2+3}\) = \(\frac{-6+21}{5}\) = \(\frac{15}{5}\) = 3

Therefore, the required point is (1, 3).

Question 2.

Find the coordinates of the points of tri-section of the line segment joining (4, -1) and (-2, -3).

Solution:

Let P and Q be the points of the tri-section of AB.

Then,

AP = PQ = QB = 1

Case I: Here P divides AB in the ratio 1 : 2.

So, we have

x_{1} = 4, y_{1} = -1

x_{2} = -2, y_{1} = -3

and m_{1} = 1, m_{2} = 2

∴ The coordinates of P are given by

= p \(\left[\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}+\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right]\)

= p \(\left[\frac{1(-2)+2(4)}{1+2}, \frac{1(-3)+2(-1)}{1+2}\right]\)

= p \(\left[\frac{-2+8}{3}, \frac{-3-2}{3}\right]\)

= p \(\left[\frac{6}{3}, \frac{-5}{3}\right]\)

= p \(\left[\frac{6}{3}, \frac{-5}{3}\right]\)

= p \(\left[2, \frac{-5}{3}\right]\)

Case II:

Here, Q divides AB in the ratio 2: 1.

So, we have

x_{1} = 4, y_{1} = -1

x_{2} = -2, y_{2} = -3

and m_{1} = 2, m_{2} = 1

∴ The coordinates of Q are given by

∴ Coordinates of required points are \(\left[2, \frac{-5}{3}\right]\) and \(\left[0, \frac{-7}{3}\right]\).

Question 3.

To conduct Sports Day activities in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Figure. Niharika runs \(\frac {1}{4}\)th the distance AD on the 2 line and posts a green flag. Preet runs \(\frac {1}{5}\)th distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

From the above figure, the position of given flag posted Niharika is M(2 \(\frac {1}{4}\) x 100), i.e. M(2, 25)

and red flag post by Preet is N(8, \(\frac {1}{5}\) x 100), i.e., N(8, 20).

Now, MN = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)

= \(\sqrt{6^{2}+(-5)^{2}}\)

= \(\sqrt{36+25}\) = \(\sqrt{61}\)

Hence, the distance between the flags = \(\sqrt{61}\) m and position of the blue flag post be

P is given by = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)

= \(\left(\frac{10}{2}, \frac{45}{2}\right)\) = (5, 22.5)

Hence, the blue flag is on the fifth line at a distance of 22.5 m above it.

Question 4.

Find the ratio in which the segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Solution:

Let P(-1, 6) divides the line segment joining the points A(-3, 10) and B(6, -8) in the ratio k: 1.

Then the coordinates of P are given by –

Here, we have

x_{1} = -3, y_{1} = 10

x_{2} = 6, y_{2} = -8

and m_{1} = k, m_{2} = 1

So, the coordinates of P are

But the coordinates of P are given as P(-1, 6)

∴ \(\frac{6 k-3}{k+1}\) = -1

⇒ 6k -3 = -1(k + 1)

⇒ 6k -3 = -k – 1

⇒ 7k = 2

⇒ k = \(\frac{2}{7}\)

and = \(\frac{2}{7}\) = 6

⇒ -8k + 10 = 6(k + 1)

⇒ -8k + 10 = 6k + 6

⇒ -8k – 6k = 6 – 10

⇒ -14k = -4

⇒ k = \(\frac{4}{14}\) = \(\frac{2}{7}\)

Hence, required ratio = 2 : 7.

Question 5.

Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of divisions.

Solution:

Let the given points be A(1, -5) and B(-4, 5).

Let the x-axis cuts AB at the point P in the ratio k :1.

Then, the coordinates of P are given as

Here, we have

x_{1} = 1, y_{1} = -5

x_{2} = -4 y_{2} = 5

and m_{1} = k, m_{1} = 1

So, coordinates of P are

Since, P lies on x-axis. So its ordinate is 0.

∴ \(\frac{5 k-5}{k+1}\) = 0

⇒ 5k – 5 = 0

⇒ 5k = 5

⇒ k = 1

Hence, required ratio = (1 : 1).

Now, the coordinates of the point of division are

Question 6.

If(1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.

We know that the diagonals of parallelogram bisect each other. So,

Coordinates of the mid-point of AC

= Coordinates of the mid-point of BD

⇒ \(\left(\frac{1+x}{2}, \frac{2+6}{2}\right)\) = \(\left(\frac{4+3}{2}, \frac{y+5}{2}\right)\)

⇒ \(\left(\frac{1+x}{2}, 4\right)\) = \(\left(\frac{7}{2}, \frac{y+5}{2}\right)\)

⇒ \(\frac {1 + x}{2}\) = \(\frac {7}{2}\)

⇒ 1 + x = 7

⇒ x = 6

and y = \(\frac {y + 5}{2}\)

⇒ y + 5 = 8

⇒ y = 3

Hence, required values are x = 6 and y = 3.

Question 7.

Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, -3) and B is (1, 4).

Solution:

Let the given point be A(x, y). Since C is the mid-point of AB.

∴ The coordinates of C are

= C \(\left[\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right]\)

= C \(\left[\frac{x+1}{2}, \frac{y+4}{2}\right]\)

But, the coordinates of C are (2, -3)

⇒ \(\frac {1 + x}{2}\) = 2 and \(\frac {y + 4}{2}\) = -3

⇒ x + 1 = 4 and y + 4 = -6

x = 4 – 1 and y = -6 -4

x = 3 and y = -10

Hence, the coordinates of A are (3, -10).

Question 8.

If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = AB and P lies on the line segment AB.

Solution:

AP = \(\frac {3}{7}\)AB

7AP = 3AB

7AP = 3(AP + PB)

[∴ p lines in the line segment AB]

7AP = 3AP + 3PB

4AP = 3PB

\(\frac {AP}{PB}\) = \(\frac {3}{4}\)

We have

Let the coordinates of P be (x, y). Then,

x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)

= \(\frac{(3)(2)+(4)(-2)}{3+4}\)

= \(\frac {6 – 8}{7}\) = \(\frac {2}{7}\)

and y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)

= \(\frac{(3)(-4)+(4)(-2)}{3+4}\)

= \(\frac{-12-8}{7}\) = –\(\frac{20}{7}\)

Hence, the coordinates of the point P be (-\(\frac{2}{7}\),- \(\frac{20}{7}\))

Question 9.

Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution:

Let P, Q and R be the three points that divide the line segment joi fling the points A(-2, 2) and (2, 8) in four equal parts.

Case I: For point P, we have

Here, m_{1} = 1, m_{2} = 3

x_{1} = -2, y_{2} = 2

x_{2} = 2,y_{2} = 8

Then, coordinates of P are given by

Case II: For point Q, we have

m_{1} = 2, m_{2} = 2

x_{1} = -2, y_{2} = 2

and x_{1} = 2, y_{2} = 8

Then, coordinates of Q are given by

= Q \(\frac{-12-8}{7}\) = Q (0, 5)

Case III: For point R, we have

Question 10.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Solution:

Let the given points are A(3, 0), B(4, 5), C(-1, 4)

and D(-2, -1).

We have,

BD = \(\sqrt{(-2-4)^{2}+(-1-5)^{2}}\)

= \(\sqrt{(-6)^{2}+(-6)^{2}}\)

= \(\sqrt{36+36}\) = \(\sqrt{72}\)

AC = \(\sqrt{(-1-3)^{2}+(4-0)^{2}}\)

= \(\sqrt{(-4)^{2}+(4)^{2}}\)

= \(\sqrt{16+16}\) = \(\sqrt{32}\)

= 4\(\sqrt{2}\)

Now, area of rhombus ABCD

= \(\frac{1}{2}\) x (Product of diagonals)

Area of rhombus = x (AC x BD)

= \(\left[\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}\right]\) sq. units

= (2\(\sqrt{2}\) x 6\(\sqrt{2}\)) sq. units

= 24 sq. units